Strict concavity in the closure of a convex set

Question

Is the following statement true?

Let $f:X\to \mathbb R$, $X$ normed vector space, be a continuous function that is strictly concave function over a convex set $C$, then $f$ is strictly concave over $\bar C$ (the closure of $C$).

How about something like $x^2+xy^2$ and $C = (0,1) \times (0,1)$? — Matias Heikkilä, Aug 22 '20 at 20:14

score 1 · Accepted Answer · answered Aug 23 '20 at 14:21

1

Hint: try the counterexample $f(x,y)=\sqrt[3]{xy}$ on $x>0$, $y>0$.

answered Aug 23 '20 at 14:21

A.Γ.

29,518

You are right. Your counterexample works. Thanks. – Condor5 Aug 23 '20 at 18:03

score 0 · Answer 2 · answered Aug 23 '20 at 13:50

0

I don't think that statement is correct. Consider the function $$f(x)= \begin{cases} &-x^2 & ,x>0 \\ &1 &,x=0 \end{cases}$$ and let the set $C =\{ x\,|\, x>0\} $. Then $f$ is strictly concave over $C$, but it is not concave over $Cl (C)$.

answered Aug 23 '20 at 13:50

iarbel84

1,513

But this function is not continuous. – Condor5 Aug 23 '20 at 17:32
I see my statement was ambiguous. I have edited it. – Condor5 Aug 23 '20 at 17:38

Strict concavity in the closure of a convex set

2 Answers2